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ilikephysics2
Group Title
Find the first 4 terms of the taylor series of sin x at x = pi/4.
 one year ago
 one year ago
ilikephysics2 Group Title
Find the first 4 terms of the taylor series of sin x at x = pi/4.
 one year ago
 one year ago

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ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
The first 4 derivatives are cos(x) then sin(x) then cos(x) and then sin(x)
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
sin(pi/4) + cos(pi/4) * (xpi/4) sin(pi/4) * (1/2!) (xpi/4)^2 .....
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
is that it?
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
pretty straightforward.. what was your question about?
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
that's the first 3 terms... you should be able to get the 4th now?
 one year ago

ilikephysics2 Group TitleBest ResponseYou've already chosen the best response.0
That does not look right though
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
well... I lied a bit... a few of those terms are zero...
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.0
but you can work from them... right?
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.0
f(x) = f(x0) + (xx0)*f'(x0)/1! + (xx0)^2*f"(x0)/2! + (xx0)^3*f'"(x0)/3!
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.0
with x0 = pi/4, for f(x) = sinx > f(pi/4) = sin(pi/4) = 1/2*sqrt(2) f'(x) = cosx > f'(pi/4) = cos(pi/4) = 1/2*sqrt(2) f"(x) = sinx > f"(pi/4) = sin(pi/4) = 1/2*sqrt(2) f'"(x) = cosx > f'"(pi/4) = cos(pi/4) = 1/2*sqrt(2) now, just subtitute all numbers above to generally formula !!!
 one year ago
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